- #36

PeterDonis

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Just look at them. But first you have to get them right.Buzz Bloom said:I do not understand why the integrand I uses is not equivalent to the integrand you posted.

Yes, that's what you posted earlier.Buzz Bloom said:Mine is

$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.

No, it isn't. I think you mistakenly quoted yourself again. My integrand is this:Buzz Bloom said:Yours is based on

$$dt = \frac {dx} {3H_0 \sqrt L \sqrt{x^2 + 2Qx} }$$.

These are obviously not the same function. The integration variable being ##x## vs. ##a## is a cosmetic detail, the name of the variable doesn't matter, what matters is the functional form of the integrand. Since in both cases, the range of integration is ##0## to ##1##, and the functions being integrated are different, obviously the numerical answers will be different. And, as you have already found, your integrand gives the wrong answer. Mine gives the right answer.PeterDonis said:$$dt = \frac{1}{H_0} \frac{da}{\sqrt{\frac{M}{a} + L a^2}}$$

Evidently you made some mistake in the substitutions you made to get your integrand in ##x## from the original one. That's why you're getting the wrong answer.